The number of all 3-digit numbers abc for which abc + ab+bc + ac+a+b+c = 29

$$(a+1)(b+1)(c+1)=30$$

Total number of ways to satisfy above equation:

1. $30 = (2,3,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=1, b=2, c=4)$
2. $30 = ( 1 , 6 ,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=0, b=5, c=4)$
3. $30 = (1,3,10)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=0, b=2, c=9)$

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In the first case we have $6$ values for $a,b,c$ as $(1,2,4)$ can take up any digit hence they can be arranged in $3!$ ways.

In the second case we have $4$ values for $a,b,c$ as $(0,5,4)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .

Similarly in the third case we have $4$ values $a,b,c$ as $(0,2,9)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .

Therefore, total number of ways:

$$3!+2\times 2\times1+2\times 2\times1=6+4+4=14 \space \text{ways}$$

The prime fctorisation of $30$ is $5\times3\times2$. The only ordered 3-uples for $a+1,b+1$ and $c+1$ where $a+1 \in [2,10]$ and $b+1,c+1 \in [1,10]$ are : $$(2,3,5),(2,5,3),(3,1,10),(3,2,5),(3,5,2),(3,10,1),(5,1,6),(5,2,3),(5,3,2),(5,6,1),(6,1,5),(6,5,1),(10,1,3),(10,3,1)$$

So the solutions are : $$124,142,209,214,241,290,405,412,421,450,504,540,902,920$$

A quick bruteforce approach to figuring out the possible number of solutions:

Javascript code, paste in browser console for a quick test.

for(i=100;i<=999;i++){
    a=i.toString();
    //Take the individual characters(the digits) of the string, increment by 1 and multiply.
    if(++a[0]*++a[1]*++a[2] === 30){
        console.log(i);
    }
}

This gives the following 14 numbers:
124 142 209 214 241 290 405 412 421 450 504 540 902 920